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Let $u,v \in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?

Studying Optimization
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3 Answers3

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Assuming $uv^*\ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $\det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields $$ \det(uv^*+I)=v^*u+1 $$

egreg
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    Very clever short proof. – Jean Marie Dec 09 '18 at 23:27
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    @JeanMarie And the general formula for $\det(A+uv^)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^$ (which has rank at most $1$) relative to $v^*A^{-1}u$. – egreg Dec 09 '18 at 23:47

  • "Since the matrix has rank $1$, the characteristic polynomial is $\det(uv^-XI)=(v^u-X)(0-X)^{n-1}$."

    Why is this true?

     – John D Nov 09 '23 at 05:24
  • @JohnD The geometric multiplicity of $0$ is $n-1$, so the algebraic multiplicity must be at least $n-1$, but it cannot be greater because of the nonzero eigenvalue, so it is $n-1$. – egreg Nov 09 '23 at 08:48
  • I understand uv* is rank 1 and that XI is full rank, how do we know we can say anything about their difference? And how do we know their difference has eigenvalues v*u - X and -X? Why do we have linearity in the eigenvalues of this new difference matrix? – John D Nov 09 '23 at 20:01
  • @JohnD You're looking from the wrong perspective. By definition, the characteristic polynomial of a matrix $A$ is $\det(A-XI)$, where $X$ is an indeterminate considered as a scalar. We know that we can factor the characteristic polynomial as $(\lambda_1-X)^{m_1}\dotsm(\lambda_r-X)^{m_r}$, where $\lambda_1,\dots,\lambda_r$ are the distinct eigenvalues (in the complex numbers). The dimension $d_k$ of the eigenspace relative to $\lambda_k$ satisfies $d_k\le m_k$. Here $r=2$, $\lambda_1=v^*u$ with $m_1\ge1$ and $\lambda_2=0$ with $d_2=n-1$. The only possibility for $m_2$ is $n-1$, and $m_1=1$. – egreg Nov 09 '23 at 20:22
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Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that $$ {\displaystyle \det \left({A} +{uv} ^{\textsf {T}}\right) =\left(1+{v} ^{\textsf {T}}{A} ^{-1}{u} \right)\,\det \left({A} \right)\,.} $$

In your case, let $A=I$.

The proof is essentially based on the following observation: $$ {\displaystyle {\begin{pmatrix}\mathbf {I} &0\\\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} +\mathbf {uv} ^{\textsf {T}}&\mathbf {u} \\0&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} &0\\-\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}={\begin{pmatrix}\mathbf {I} &\mathbf {u} \\0&1+\mathbf {v} ^{\textsf {T}}\mathbf {u} \end{pmatrix}}.} $$

[Added later:] Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{\textsf {T}}$ is a rank one matrix. Moreover, $\lambda = {v} ^{\textsf {T}}u$ is an eigenvalue of $B$ since: $$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal: $$ (v^Tu,0,\cdots,0). $$ It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal: $$ (1+v^Tu,1,\cdots,1). $$ But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.

1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$

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You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u\,.$$

Fabian
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